16.333: Lecture #1
Equilibrium States
Aircraft performance
Introduction to basic terms
Fall 2004
16.333 1–1
Aircraft Performance
• Accelerated horizontal flight ­ balance of forces
– Engine thrust T
L
– Lift L (⊥ to V )
– Drag D (� to V )
D
T
– Weight W
mg
T −D =m
dV
= 0 for steady flight
dt
and
L−W =0
• Define L = 21 ρV 2SCL where
ct
– ρ – air density (standard tables)
– S – gross wing area = c̄ × b,
c
Trailing
edge
– c̄ = mean chord
c0
b = 2.s
– b = wing span
– AR – wing aspect ratio = b/c̄
Sweepback angle �k
kc
– Q = 12 ρV 2 dynamic pressure
– V = speed relative to the air
Fall 2004
16.333 1–2
– CL lift coefficient – for low Mach number, CL = CLα (α − α0)
3 α angle of incidence of wind to the wing
3 α0 is the angle associated with zero lift
• Back to the performance:
1
L = ρV 2SCL and L = mg
2
�
2mg
which implies that V = ρSC
so that
L
−1/2
V ∝ CL
and we can relate the effect of speed to wing lift
• A key number is stall speed, which is the lowest speed that an aircraft
can fly steadily
�
2mg
Vs =
ρSCLmax
where typically get CLmax at αmax = 10◦
Fall 2004
16.333 1–3
Steady Gliding Flight
• Aircraft at a steady glide angle of γ
• Assume forces are in equilibrium
L − mg cos γ = 0
D + mg sin γ = 0
Gives that
D CD
≡
L
CL
⇒ Minimum gliding angle obtained when CD /CL is a minimum
tan γ =
– High L/D gives a low gliding angle
• Note: typically
CL2
CD = CDmin +
πARe
where
– CDmin is the zero lift (friction/parasitic) drag
2
– CL
gives the lift induced drag
– e is Oswald’s efficiency factor ≈ 0.7 − 0.85
(1)
(2)
Fall 2004
16.333 1–4
• Total drag then given by
�
1 2
1 2 �
2
D = ρV SCD = ρV S CDmin + kCL
2
2
1 2
(mg)2
= ρV SCDmin + k 1 2
2
2 ρV S
D
Total
drag
No-lift drag
Lift-dependent
drag
VS
1
VEmd
VE
• So that the speed for minimum drag is
�
�
�1/4
k
2mg
Vmin drag =
ρS CDmin
(3)
(4)
Fall 2004
16.333 1–5
Steady Climb
T-D
L
V
�
T
D
�
R/C
W
• Equations:
T − D − W sin γ = 0
L − W cos γ = 0
� which gives
T −D−
(5)
(6)
L
cos γ = 0
sin γ
so that
tan γ =
T −D
L
• Consistent with 1–3 if T = 0 since then γ as defined above is negative
• Note that for small γ, tan γ ≈ γ ≈ sin γ
(T − D)V
L
so that the rate of climb is approximately equal to the excess power
available (above that needed to maintain level flight)
R/C = V sin γ ≈ V γ ≈
Fall 2004
16.333 1–6
Steady Turn
L sin �
L cos �
L
�
Centrifugal
force
R
Radius of turn
�
W
• Equations:
L sin φ = centrifugal force
mV 2
=
R
L cos φ = W = mg
V2
⇒ tan φ =
Rg
V =Rω
=
(8)
(9)
Vω
g
• Note: obtain Rmin at CLmax
WV 2
1 2
Rmin( ρV SCLmax ) sin φ =
2
g
⇒ Rmin =
(7)
W/S
1/2ρgCLmax sin φmax
where W/S is the wing loading and φmax < 30◦
(10)
Fall 2004
16.333 1–7
• Define load factor N = L/mg. i.e. ratio of lift in turn to weight
N = sec φ = (1 + tan2 φ)1/2
�
tan φ =
N2 − 1
so that
V2
V2
R=
= √
g tan φ g N 2 − 1
• For a given load factor (wing strength)
R ∝V2
• Compare straight level with turning flight
– If same light coefficient
L
mg
1
CL = 1 2 = 1 2 ≡ N mg ρVt2S
2
2 ρV S
2 ρV S
√
so that Vt = N V gives the speed increase (more lift)
• Note that CL constant ⇒ CD constant ⇒ D ∝ V 2CD
⇒ Tt ∝ Dt ∝ Vt2CD ∼ N D
so that must increase throttle or will descend in the turn
(11)
(12)